QBManager

ELECTROCHEMISTRY

276096 The emf of the Daniel Cell $Zn | {ZnSO}_{4} (0.01 M) | | {CuSO}_{4} (1 M) | Cu$ at $298 K$ is $E_{1}$ . When concentration of ${ZnSO}_{4}$ is changed to $1 M$ and that of ${CuSO}_{4}$ is changed to $0.01 M$ . the emf changed to $E_{2}$ . Then find the relationship between $E_{1}$ and $E_{2}$ .

1

E_{1} > E_{2}

2

E_{1} < E_{2}

3

E_{1} = E_{2}

4

E_{2} = 0 \neq E_{1}

Explanation:

According to Nernst equation:-
$\begin{aligned} E_{cell} = E_{cell}^{o} - \frac{0.0591}{n} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ = 1.1 - \frac{0.0591}{2} \log \frac{[0.01]}{1} \\ = 1.1 - \frac{0.0591}{2} \times \log [1 \times 10^{- 2}] \\ = 1.1 + \frac{0.0591}{2} \times 2 \\ = 1.16 V \end{aligned}$
Again, when concentration changed,
$E_{cell} = E_{cell}^{o} - \frac{0.0591}{2} \log [\frac{1}{0.01}]$
$E_{cell} = 1.16 - \frac{0.0591}{2} \log 10^{2}$
$E_{cell} = 1.16 - 0.0591$
$E_{cell} = 1.041 V$
So, $E_{1} > E_{2}$

Shift-II

ELECTROCHEMISTRY

276097 Calculate the maximum work that can be obtained from the cell,
$Z n | {Z n}^{2 +} (1 M) | | {A g}^{+} (1 M) | A g$
Where $E^{\circ}_{{Zn}^{2 +} ∣ Zn} = - 0.76 V$ and $E^{\circ}_{{Ag}^{+} ∣ Ag} = 0.80 V$

1

- 301.080 kJ

2

201.830 kJ

3

112.830 kJ

4

212.630 kJ

Explanation:

$\begin{aligned} E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} \\ = 0.80 - (- 0.76) \\ = + 1.56 V \\ Zn | {Zn}^{2 +} (1 M) | {Ag}^{+} (1 M) | Ag \\ Zn + e^{-} ⟶ {Zn}^{-} \\ W_{max} = Δ G^{\circ} = - {nFE}^{\circ} \\ = - 2 \times 96500 \times 1.56 \\ = - 301080.00 J \\ = - 301.080 kJ \end{aligned}$

AMU-2013

ELECTROCHEMISTRY

276098 For a cell reaction involving a two electron change, the standard emf of the cell is found to be $0.295 V$ at $25^{\circ} C$ . The equilibrium constant of the reaction at $25^{\circ} C$ will be

1

1 \times 10^{- 10}

2

29.5 \times 10^{- 2}

3 10

4

1 \times 10^{10}

Explanation:

$E_{cell}^{\circ} = \frac{0.059}{n} \log K$
$\begin{aligned} ∴ 0.295 = \frac{0.059}{n} \log K \\ \log K = \frac{0.295 \times 2}{0.059} = 10 \\ K = 1 \times 10^{10} \end{aligned}$

COMEDK-2010

ELECTROCHEMISTRY

276099 Given the data at $25^{\circ} C$
$Ag + I^{-} \to AgI + e^{-}; E^{0} = 0.152 V$
$Ag \to {Ag}^{+} + e^{-}; E^{o} = - 0.8 V$
What is the value of $\log K_{sp}$ for AgI?
$(2.303 \frac{RT}{F} = 0.059 V)$

1 -8.12

2 +8.612

3 -37.83

4 -16.13

Explanation:

$Ag ⟶ {Ag}^{+} + e^{-}; E^{\circ} = - 0.8 V$
$Ag ⟶ {Ag}^{+} + I; E^{\circ} = - 0.952$
$E_{cell}^{\circ} = \frac{0.059}{n} \log K_{sp}$
$- 0.952 = \frac{0.0952}{1} \log K_{sp}$
$\log K_{sp} = \frac{0.952}{0.059}$
$= - 16.135$

AIEEE 2006

ELECTROCHEMISTRY

276096 The emf of the Daniel Cell $Zn | {ZnSO}_{4} (0.01 M) | | {CuSO}_{4} (1 M) | Cu$ at $298 K$ is $E_{1}$ . When concentration of ${ZnSO}_{4}$ is changed to $1 M$ and that of ${CuSO}_{4}$ is changed to $0.01 M$ . the emf changed to $E_{2}$ . Then find the relationship between $E_{1}$ and $E_{2}$ .

1

E_{1} > E_{2}

2

E_{1} < E_{2}

3

E_{1} = E_{2}

4

E_{2} = 0 \neq E_{1}

Explanation:

According to Nernst equation:-
$\begin{aligned} E_{cell} = E_{cell}^{o} - \frac{0.0591}{n} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ = 1.1 - \frac{0.0591}{2} \log \frac{[0.01]}{1} \\ = 1.1 - \frac{0.0591}{2} \times \log [1 \times 10^{- 2}] \\ = 1.1 + \frac{0.0591}{2} \times 2 \\ = 1.16 V \end{aligned}$
Again, when concentration changed,
$E_{cell} = E_{cell}^{o} - \frac{0.0591}{2} \log [\frac{1}{0.01}]$
$E_{cell} = 1.16 - \frac{0.0591}{2} \log 10^{2}$
$E_{cell} = 1.16 - 0.0591$
$E_{cell} = 1.041 V$
So, $E_{1} > E_{2}$

Shift-II

ELECTROCHEMISTRY

276097 Calculate the maximum work that can be obtained from the cell,
$Z n | {Z n}^{2 +} (1 M) | | {A g}^{+} (1 M) | A g$
Where $E^{\circ}_{{Zn}^{2 +} ∣ Zn} = - 0.76 V$ and $E^{\circ}_{{Ag}^{+} ∣ Ag} = 0.80 V$

1

- 301.080 kJ

2

201.830 kJ

3

112.830 kJ

4

212.630 kJ

Explanation:

$\begin{aligned} E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} \\ = 0.80 - (- 0.76) \\ = + 1.56 V \\ Zn | {Zn}^{2 +} (1 M) | {Ag}^{+} (1 M) | Ag \\ Zn + e^{-} ⟶ {Zn}^{-} \\ W_{max} = Δ G^{\circ} = - {nFE}^{\circ} \\ = - 2 \times 96500 \times 1.56 \\ = - 301080.00 J \\ = - 301.080 kJ \end{aligned}$

AMU-2013

ELECTROCHEMISTRY

276098 For a cell reaction involving a two electron change, the standard emf of the cell is found to be $0.295 V$ at $25^{\circ} C$ . The equilibrium constant of the reaction at $25^{\circ} C$ will be

1

1 \times 10^{- 10}

2

29.5 \times 10^{- 2}

3 10

4

1 \times 10^{10}

Explanation:

$E_{cell}^{\circ} = \frac{0.059}{n} \log K$
$\begin{aligned} ∴ 0.295 = \frac{0.059}{n} \log K \\ \log K = \frac{0.295 \times 2}{0.059} = 10 \\ K = 1 \times 10^{10} \end{aligned}$

COMEDK-2010

ELECTROCHEMISTRY

276099 Given the data at $25^{\circ} C$
$Ag + I^{-} \to AgI + e^{-}; E^{0} = 0.152 V$
$Ag \to {Ag}^{+} + e^{-}; E^{o} = - 0.8 V$
What is the value of $\log K_{sp}$ for AgI?
$(2.303 \frac{RT}{F} = 0.059 V)$

1 -8.12

2 +8.612

3 -37.83

4 -16.13

Explanation:

$Ag ⟶ {Ag}^{+} + e^{-}; E^{\circ} = - 0.8 V$
$Ag ⟶ {Ag}^{+} + I; E^{\circ} = - 0.952$
$E_{cell}^{\circ} = \frac{0.059}{n} \log K_{sp}$
$- 0.952 = \frac{0.0952}{1} \log K_{sp}$
$\log K_{sp} = \frac{0.952}{0.059}$
$= - 16.135$

AIEEE 2006

ELECTROCHEMISTRY

276096 The emf of the Daniel Cell $Zn | {ZnSO}_{4} (0.01 M) | | {CuSO}_{4} (1 M) | Cu$ at $298 K$ is $E_{1}$ . When concentration of ${ZnSO}_{4}$ is changed to $1 M$ and that of ${CuSO}_{4}$ is changed to $0.01 M$ . the emf changed to $E_{2}$ . Then find the relationship between $E_{1}$ and $E_{2}$ .

1

E_{1} > E_{2}

2

E_{1} < E_{2}

3

E_{1} = E_{2}

4

E_{2} = 0 \neq E_{1}

Explanation:

According to Nernst equation:-
$\begin{aligned} E_{cell} = E_{cell}^{o} - \frac{0.0591}{n} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ = 1.1 - \frac{0.0591}{2} \log \frac{[0.01]}{1} \\ = 1.1 - \frac{0.0591}{2} \times \log [1 \times 10^{- 2}] \\ = 1.1 + \frac{0.0591}{2} \times 2 \\ = 1.16 V \end{aligned}$
Again, when concentration changed,
$E_{cell} = E_{cell}^{o} - \frac{0.0591}{2} \log [\frac{1}{0.01}]$
$E_{cell} = 1.16 - \frac{0.0591}{2} \log 10^{2}$
$E_{cell} = 1.16 - 0.0591$
$E_{cell} = 1.041 V$
So, $E_{1} > E_{2}$

Shift-II

ELECTROCHEMISTRY

276097 Calculate the maximum work that can be obtained from the cell,
$Z n | {Z n}^{2 +} (1 M) | | {A g}^{+} (1 M) | A g$
Where $E^{\circ}_{{Zn}^{2 +} ∣ Zn} = - 0.76 V$ and $E^{\circ}_{{Ag}^{+} ∣ Ag} = 0.80 V$

1

- 301.080 kJ

2

201.830 kJ

3

112.830 kJ

4

212.630 kJ

Explanation:

$\begin{aligned} E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} \\ = 0.80 - (- 0.76) \\ = + 1.56 V \\ Zn | {Zn}^{2 +} (1 M) | {Ag}^{+} (1 M) | Ag \\ Zn + e^{-} ⟶ {Zn}^{-} \\ W_{max} = Δ G^{\circ} = - {nFE}^{\circ} \\ = - 2 \times 96500 \times 1.56 \\ = - 301080.00 J \\ = - 301.080 kJ \end{aligned}$

AMU-2013

ELECTROCHEMISTRY

276098 For a cell reaction involving a two electron change, the standard emf of the cell is found to be $0.295 V$ at $25^{\circ} C$ . The equilibrium constant of the reaction at $25^{\circ} C$ will be

1

1 \times 10^{- 10}

2

29.5 \times 10^{- 2}

3 10

4

1 \times 10^{10}

Explanation:

$E_{cell}^{\circ} = \frac{0.059}{n} \log K$
$\begin{aligned} ∴ 0.295 = \frac{0.059}{n} \log K \\ \log K = \frac{0.295 \times 2}{0.059} = 10 \\ K = 1 \times 10^{10} \end{aligned}$

COMEDK-2010

ELECTROCHEMISTRY

276099 Given the data at $25^{\circ} C$
$Ag + I^{-} \to AgI + e^{-}; E^{0} = 0.152 V$
$Ag \to {Ag}^{+} + e^{-}; E^{o} = - 0.8 V$
What is the value of $\log K_{sp}$ for AgI?
$(2.303 \frac{RT}{F} = 0.059 V)$

1 -8.12

2 +8.612

3 -37.83

4 -16.13

Explanation:

$Ag ⟶ {Ag}^{+} + e^{-}; E^{\circ} = - 0.8 V$
$Ag ⟶ {Ag}^{+} + I; E^{\circ} = - 0.952$
$E_{cell}^{\circ} = \frac{0.059}{n} \log K_{sp}$
$- 0.952 = \frac{0.0952}{1} \log K_{sp}$
$\log K_{sp} = \frac{0.952}{0.059}$
$= - 16.135$

AIEEE 2006

ELECTROCHEMISTRY

276096 The emf of the Daniel Cell $Zn | {ZnSO}_{4} (0.01 M) | | {CuSO}_{4} (1 M) | Cu$ at $298 K$ is $E_{1}$ . When concentration of ${ZnSO}_{4}$ is changed to $1 M$ and that of ${CuSO}_{4}$ is changed to $0.01 M$ . the emf changed to $E_{2}$ . Then find the relationship between $E_{1}$ and $E_{2}$ .

1

E_{1} > E_{2}

2

E_{1} < E_{2}

3

E_{1} = E_{2}

4

E_{2} = 0 \neq E_{1}

Explanation:

According to Nernst equation:-
$\begin{aligned} E_{cell} = E_{cell}^{o} - \frac{0.0591}{n} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ = 1.1 - \frac{0.0591}{2} \log \frac{[0.01]}{1} \\ = 1.1 - \frac{0.0591}{2} \times \log [1 \times 10^{- 2}] \\ = 1.1 + \frac{0.0591}{2} \times 2 \\ = 1.16 V \end{aligned}$
Again, when concentration changed,
$E_{cell} = E_{cell}^{o} - \frac{0.0591}{2} \log [\frac{1}{0.01}]$
$E_{cell} = 1.16 - \frac{0.0591}{2} \log 10^{2}$
$E_{cell} = 1.16 - 0.0591$
$E_{cell} = 1.041 V$
So, $E_{1} > E_{2}$

Shift-II

ELECTROCHEMISTRY

276097 Calculate the maximum work that can be obtained from the cell,
$Z n | {Z n}^{2 +} (1 M) | | {A g}^{+} (1 M) | A g$
Where $E^{\circ}_{{Zn}^{2 +} ∣ Zn} = - 0.76 V$ and $E^{\circ}_{{Ag}^{+} ∣ Ag} = 0.80 V$

1

- 301.080 kJ

2

201.830 kJ

3

112.830 kJ

4

212.630 kJ

Explanation:

$\begin{aligned} E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} \\ = 0.80 - (- 0.76) \\ = + 1.56 V \\ Zn | {Zn}^{2 +} (1 M) | {Ag}^{+} (1 M) | Ag \\ Zn + e^{-} ⟶ {Zn}^{-} \\ W_{max} = Δ G^{\circ} = - {nFE}^{\circ} \\ = - 2 \times 96500 \times 1.56 \\ = - 301080.00 J \\ = - 301.080 kJ \end{aligned}$

AMU-2013

ELECTROCHEMISTRY

276098 For a cell reaction involving a two electron change, the standard emf of the cell is found to be $0.295 V$ at $25^{\circ} C$ . The equilibrium constant of the reaction at $25^{\circ} C$ will be

1

1 \times 10^{- 10}

2

29.5 \times 10^{- 2}

3 10

4

1 \times 10^{10}

Explanation:

$E_{cell}^{\circ} = \frac{0.059}{n} \log K$
$\begin{aligned} ∴ 0.295 = \frac{0.059}{n} \log K \\ \log K = \frac{0.295 \times 2}{0.059} = 10 \\ K = 1 \times 10^{10} \end{aligned}$

COMEDK-2010

ELECTROCHEMISTRY

276099 Given the data at $25^{\circ} C$
$Ag + I^{-} \to AgI + e^{-}; E^{0} = 0.152 V$
$Ag \to {Ag}^{+} + e^{-}; E^{o} = - 0.8 V$
What is the value of $\log K_{sp}$ for AgI?
$(2.303 \frac{RT}{F} = 0.059 V)$

1 -8.12

2 +8.612

3 -37.83

4 -16.13

Explanation:

$Ag ⟶ {Ag}^{+} + e^{-}; E^{\circ} = - 0.8 V$
$Ag ⟶ {Ag}^{+} + I; E^{\circ} = - 0.952$
$E_{cell}^{\circ} = \frac{0.059}{n} \log K_{sp}$
$- 0.952 = \frac{0.0952}{1} \log K_{sp}$
$\log K_{sp} = \frac{0.952}{0.059}$
$= - 16.135$

AIEEE 2006

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